Let $S$ be a finite, cancellative semigroup under a binary operation $ \theta $. Then $(S,\theta)$ is a group.

Question

Let $S$ be a finite, cancellative semigroup under a binary operation $ \theta $. Then $(S,\theta)$ is a group.

The proof of this statement is already given here, but I would like to check if this is an equivalence relationship? I.e. I want to check if the reciprocal is also a true statement?

No, because there exist groups which are not finite. Though it is true that every group is a cancellitive semigroup — Chris Grossack, Dec 09 '19 at 22:52
Yes. Every finite group is a finite cancellitive semigroup. This is basically because groups are semigroups with "bonus properties". — Chris Grossack, Dec 09 '19 at 22:54

score 1 · Accepted Answer · answered Dec 09 '19 at 22:55

1

Well, no: there are groups that are not finite. If you are asking if all finite groups are cancellative, then yes (and, indeed, all groups are cancellative): if $G$ is a group and $a, b, c \in G$ such that $ab = ac$, we have $b = 1b = (a^{-1}a)b = a^{-1}(ab) = a^{-1}(ac) = (a^{-1}a)c = 1c = c$, so $G$ is cancellative (where those equalities are the identity, inverse, and associativity axioms of a group, then the assumption that $ab = ac$, and finally those same axioms in reverse order).

answered Dec 09 '19 at 22:55

user3482749

6,750

So if you have a finite group, is it necessarily a finite, cancellative semigroup? – Dec 09 '19 at 22:57
Yes. You can also drop the two "finite"s. – user3482749 Dec 09 '19 at 22:58

Let $S$ be a finite, cancellative semigroup under a binary operation $ \theta $. Then $(S,\theta)$ is a group.

1 Answers1